A fourth-order tensor Tijkl has the properties Tjikl = −Tijkl, Tijlk = −Tijkl. Prove that for any such tensor there exists a second-order tensor Kmn such that Tijkl = εijmεklnKmn and give an explicit expression for Kmn. Consider two (separate) special cases, as follows. (a) Given that Tijkl is

Question

A fourth-order tensor [latex]T_{i j k l}[/latex] has the properties

[latex]T_{j i k l}=-T_{i j k l}, \quad T_{i j l k}=-T_{i j k l}.[/latex]

Prove that for any such tensor there exists a second-order tensor [latex]K_{m n}[/latex] such that

[latex]T_{i j k l}=\epsilon_{i j m} \epsilon_{k l n} K_{m n}[/latex]

and give an explicit expression for [latex]K_{m n}[/latex]. Consider two (separate) special cases, as follows.

(a) Given that [latex]T_{i j k l}[/latex] is isotropic and [latex]T_{i j j i}[/latex] = 1, show that [latex]T_{i j k l}[/latex] is uniquely determined and express it in terms of Kronecker deltas.

(b) If now [latex]T_{i j k l}[/latex] has the additional property

[latex]T_{k l i j}=-T_{i j k l},[/latex]

show that [latex]T_{i j k l}[/latex] has only three linearly independent components and find an expression for [latex]T_{i j k l}[/latex] in terms of the vector

[latex]V_i=-\frac{1}{4} \epsilon_{j k l} T_{i j k l}.[/latex]

Accepted Answer

As [latex]K_{m n}[/latex] is to be a second-order tensor, we need to construct such a tensor from [latex]T_{i j k l}[/latex]. Since the latter is of fourth order, it needs to be contracted n times with a tensor of order 2n − 2 for some positive integer n. In view of the final stated expression for [latex]T_{i j k l}[/latex], involving [latex]\epsilon_{i j m} \epsilon_{k l n}[/latex], i.e. a sixth-order tensor, we try n = 4 and, starting from [latex]T_{i j k l}=\epsilon_{i j m} \epsilon_{k l n} K_{m n}[/latex], consider

[latex]\begin{aligned}\epsilon_{p i j} \epsilon_{q k l} T_{i j k l} & =\epsilon_{p i j} \epsilon_{q k l} \epsilon_{i j m} \epsilon_{k l n} K_{m n} \& =\left(\delta_{j j} \delta_{p m}-\delta_{j m} \delta_{p j} ight)\left(\delta_{l l} \delta_{q n}-\delta_{l n} \delta_{q l} ight) K_{m n} \& =\left(3 \delta_{p m}-\delta_{p m} ight)\left(3 \delta_{q n}-\delta_{q n} ight) K_{m n} \& =4 K_{p q} .\end{aligned}[/latex]

Clearly, [latex]K_{m n}=\frac{1}{4} \epsilon_{m i j} \epsilon_{n k l} T_{i j k l}[/latex] has the required property.

(a) Given that [latex]T_{i j k l}[/latex] is isotropic, and noting that [latex]\epsilon_{m i j} ext { and } \epsilon_{n k l}[/latex] are also isotropic, we conclude that [latex]K_{m n}[/latex] must itself be isotropic. It must therefore be some multiple of [latex]\delta_{m n}[/latex] (as this is the most general isotropic second-order tensor), i.e. [latex]K_{m n}=\lambda \delta_{m n}[/latex] for one or more values of [latex]\lambda[/latex]. Thus,

[latex]\begin{aligned}T_{i j k l} & =\epsilon_{i j m} \epsilon_{k \ln } \lambda \delta_{m n} \& =\lambda \epsilon_{i j m} \epsilon_{k l m} \& =\lambda\left(\delta_{i k} \delta_{j l}-\delta_{i l} \delta_{j k} ight)\end{aligned}[/latex]

Now, since [latex]T_{i j j i}[/latex] = 1,

[latex]\begin{aligned}1 & =\lambda\left(\delta_{i j} \delta_{j i}-\delta_{i i} \delta_{j j} ight) \& =\lambda\left[\delta_{i i}-\left(\delta_{i i} ight)^2 ight]=\lambda(3-9) \quad \Rightarrow \quad \lambda=-\frac{1}{6} .\end{aligned}[/latex]

We conclude that [latex]\lambda[/latex], and therefore also [latex]T_{i j k l}[/latex], is unique with [latex]T_{i j k l}=\frac{1}{6}\left(\delta_{i l} \delta_{j k}-\delta_{i k} \delta_{j l} ight)[/latex].

(b) To examine the implications of the antisymmetry indicated by [latex]T_{k l i j}=-T_{i j k l}[/latex], we interchange the pair of dummy suffices {i, j} with the pair {k, l} to obtain the third line below — and then switch them back again in the fourth line using the antisymmetry:

[latex]\begin{aligned}K_{m n} & =\frac{1}{4} \epsilon_{m i j} \epsilon_{n k l} T_{i j k l}, \K_{n m} & =\frac{1}{4} \epsilon_{n i j} \epsilon_{m k l} T_{i j k l} \& =\frac{1}{4} \epsilon_{n k l} \epsilon_{m i j} T_{k l i j} \& =-\frac{1}{4} \epsilon_{n k l} \epsilon_{m i j} T_{i j k l} \& =-K_{m n} .\end{aligned}[/latex]

Thus [latex]K_{m n}[/latex] is antisymmetric. It therefore has zeros on its leading diagonal and only three linearly independent components as non-diagonal elements. Since [latex]T_{i j k l}[/latex] is uniquely defined in terms of [latex]K_{m n}[/latex], it too has only three linearly independent components.

Now consider

[latex]\begin{aligned}\epsilon_{j k l} T_{i j k l} & =\epsilon_{j k l} \epsilon_{i j m} \epsilon_{k l n} K_{m n}, \-4 V_i & =\left(\delta_{k m} \delta_{l i}-\delta_{k i} \delta_{l m} ight) \epsilon_{k l n} K_{m n} \& =\left(\epsilon_{\min }-\epsilon_{i m n} ight) K_{m n} \& =2 \epsilon_{\min } K_{m n} .\end{aligned}[/latex]

To ‘invert’ this relationship, consider

[latex]\begin{aligned}\epsilon_{i r s} V_i & =-\frac{1}{2} \epsilon_{i r s} \epsilon_{\min } K_{m n} \& =-\frac{1}{2}\left(\delta_{r n} \delta_{s m}-\delta_{r m} \delta_{s n} ight) K_{m n} \& =-\frac{1}{2}\left(K_{s r}-K_{r s} ight) \& =K_{r s} \quad \Rightarrow \quad K_{m n}=\epsilon_{p m n} V_p .\end{aligned}[/latex]

Finally, expressing [latex]T_{i j k l}[/latex], as given in the question, explicitly in terms of the vector [latex]V_i[/latex], using the result obtained above, we have

[latex]\begin{aligned}T_{i j k l} & =\epsilon_{i j m} \epsilon_{k l n} \epsilon_{p m n} V_p \& =\left(\delta_{i n} \delta_{j p}-\delta_{i p} \delta_{j n} ight) \epsilon_{k l n} V_p \& =\epsilon_{k l i} V_j-\epsilon_{k l j} V_i.\end{aligned}[/latex]

Question 26.23: A fourth-order tensor Tijkl has the properties Tjikl = −Tijk...

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